## Initial Definition (Right Triangle) We will later expand this definition using the [[Unit Circle Definitions of Trigonometric Functions|unit circle]], but for now, given a triangle $\Delta ABC$, with $m\angle ABC = 90\textdegree$ and $m \angle CAB = \theta$, ![[Basic Trig Functions.png]] [^1] The standard mnemonic is "soh cah toa", because $\begin{align*} \sin &= \frac{\text{opposite}}{\text{hypotenuse}} \\ \cos &= \frac{\text{adjacent}}{\text{hypotenuse}} \\ \tan &= \frac{\text{opposite}}{\text{adjacent}} \\ \end{align*} $ ### Latex Note that latex has functions for `\sin`, `\cos` and `\tan`. It's also worth knowing `\theta` which is by far the most common name for an unknown angle. So you will be using `\sin \theta` etc often. You can spot at a glance if you forgot the backslash in front of a trig function because latex will think your function is text, so uses an italic font and messes up the spacing between the function and the variable. ie it will look like $sin \theta$ instead of $\sin \theta$ (right). Also although these are trig *functions*, it is common to write them without parentheses unless these are needed to disambiguate operator order. So you might often write $\sin \theta$ not $\sin (\theta)$, although the latter is not wrong. You do need parentheses for things like $\sin(90\textdegree - \theta)$ to show that you’re taking the sine of the entire angle and not just the $90\textdegree$. It is also conventional to write things like $\sin^2 \theta$ which means $(\sin \theta)^2$. See [Khan Academy Review Article](https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/intro-to-the-trig-ratios/a/finding-trig-ratios-in-right-triangles) ## Proof that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ We may notice from the above an alternative formulation for $\tan$ which will frequently be useful: $\tan \theta = \frac{\sin \theta}{\cos \theta} \tag{1}$ Let us prove this using $\Delta ABC$ in the diagram above. $ \begin{align*} \text{Given that } \sin \theta &= \frac{BC}{AC} \text{and } \cos \theta = \frac{AB}{AC} \\[8pt] \frac{\sin \theta}{\cos \theta} &= \frac{\frac{BC}{AC}}{\frac{AB}{AC}} = \frac{BC}{AC} \cdot \frac{AC}{AB} \\ &= \frac{BC}{AB}\\[8pt] \text{but } \tan \theta &= \frac{BC}{AB} \\[10pt] \text{Therefore } \tan \theta &= \frac{\sin \theta}{\cos \theta} \tag{1} \end{align*} $ ## Reciprocal trigonometric functions There are three additional trigonometric functions which can be simply derived from the above, as they are each the reciprocal of one of the basic trig functions. | Function | formula | | --------- | ----------------------------------------------------------------------------------- | | secant | $\sec \theta = \frac{1}{\cos \theta} = \frac{\text{hypotenuse}}{\text{adjacent}}$ | | cosecant | $\csc \theta = \frac{1}{\sin \theta} = \frac{\text{hypotenuse}}{\text{opposite}}$ | | cotangent | $\cot \theta = \frac{1}{\tan \theta} = \frac{\text{adjacent}}{\text{opposite}}$ | ### Avoiding confusion 1. Although the reciprocal trig functions are the multiplicative inverse, they should not be confused with the [[Inverse Trigonometric Functions]] $\arcsin$, $\arccos$ and $\arctan$ (which you will sometimes see written like that and sometimes written $\sin^{-1},\cos^{-1}, \tan^{-1}$). 2. It is very important to note that $\sec$ is the reciprocal of $\cos$ and that $\csc$ is the reciprocal of $\sin$ and not get mixed up. ## Proof that $\cot \theta = \frac{\csc \theta}{\sec \theta}$ From (1) it follows logically that $\cot \theta = \frac{\csc \theta}{\sec \theta}$, however for completeness, let's prove that independently now. $ \begin{align*} \text{Given that } \csc \theta &= \frac{AC}{BC} \text{and } \sec \theta = \frac{AC}{AB} \\[8pt] \frac{\csc \theta}{\sec \theta} &= \frac{\frac{AC}{BC}}{\frac{AC}{AB}} = \frac{AC}{BC} \cdot \frac{AB}{AC} \\ &= \frac{AB}{BC}\\[8pt] \text{but } \cot \theta &= \frac{AB}{BC} \\[10pt] \text{Therefore } \cot \theta = \frac{\csc \theta}{\sec \theta} \tag{2} \end{align*} $ Note that given (1) we could trivially derive (2) and vice versa. ## Historical trig ratios Historically, there were additional trig ratios such as *versine* ( defined I think as $\sin(1-\theta)$) and *haversine* ($\sin(1-\frac{\theta}{2}$) but these are not used any more except in specialised contexts such as navigation. [^1]: Diagram available on [geogrebra](https://www.geogebra.org/calculator/tkddwdpf)